Question

Solution of the differential equation \[ \frac{dy}{dx} - y = \cos x \] is

• A. \( y = \frac{\sin x - \cos x}{2} + ce^x \)
• B. \( y = \frac{\sin x + \cos x}{2} + ce^x \)
• C. \( y = \frac{\sin x - \cos x}{2} + ce^{-x} \)
• D. \( y = \frac{\sin x + \cos x}{2} + ce^{-x} \)

Hint:

For linear first-order differential equations, first solve the homogeneous equation and then find a particular solution. Combine both to get the general solution.

Answer 1

The Correct Option is A

We are given the differential equation: \[ \frac{dy}{dx} - y = \cos x \] Step 1: Solve the homogeneous equation.
The homogeneous equation is: \[ \frac{dy}{dx} - y = 0 \] This is a first-order linear differential equation with the solution: \[ y_h = ce^x \] Step 2: Find a particular solution.
To find a particular solution, we use the method of undetermined coefficients. We assume a solution of the form: \[ y_p = A\sin x + B\cos x \] Substitute \( y_p \) into the original differential equation: \[ \frac{d}{dx}(A\sin x + B\cos x) - (A\sin x + B\cos x) = \cos x \] \[ A\cos x - B\sin x - A\sin x - B\cos x = \cos x \] \[ - (A + B)\sin x + (A - B)\cos x = \cos x \] Equating the coefficients of \( \sin x \) and \( \cos x \): \[ A + B = 0 \quad \text{and} \quad A - B = 1 \] Solving these equations gives: \[ A = \frac{1}{2}, \quad B = -\frac{1}{2} \] Thus, the particular solution is: \[ y_p = \frac{1}{2}(\sin x - \cos x) \] Step 3: Write the general solution.
The general solution is the sum of the homogeneous and particular solutions: \[ y = y_h + y_p = ce^x + \frac{\sin x - \cos x}{2} \] Step 4: Conclusion.
Therefore, the solution to the differential equation is: \[ y = \frac{\sin x - \cos x}{2} + ce^x \] Thus, the correct answer is (A).